C++ 注释掉未使用的行后，switchcase 将无法编译

从技术上讲，SWITCH语句中的每个 case 都是一个标签.对于大约obscure and old reasons个标签，不允许将变量声明作为标签后的第一行.通过注释掉作业(job)

ptr = (struct sockaddr_in *) res->ai_addr;

线路

struct sockaddr_in *sockAddrIn = (struct sockaddr_in *) res->ai_addr;

成为标签AF_INET:后的第一行，就像我说的，在C中是非法的.

解决方案是将所有case语句放在花括号中，如下所示:

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <netdb.h>
#include <sys/types.h>
#include <sys/socket.h>
#include <arpa/inet.h>

int main (void) {

  struct addrinfo hints; 
  memset (&hints, 0, sizeof hints);

  hints.ai_family = AF_UNSPEC; 
  hints.ai_socktype = SOCK_DGRAM;  
  hints.ai_flags = AI_CANONNAME;   

  struct addrinfo *res;

  getaddrinfo ("example.com", "http", &hints, &res);
  printf ("Host: %s\n", "example.com");

  void *ptr;

  while (res != NULL) {
    printf("AI Family for current addrinfo: %i\n", res->ai_family);
    switch (res->ai_family) {
      case AF_INET:
      {
        ptr = (struct sockaddr_in *) res->ai_addr;
        struct sockaddr_in *sockAddrIn = (struct sockaddr_in *) res->ai_addr;
        break;
      }
    }
    res = res->ai_next;
  }
  return 0;
}

无论如何，我认为这是更好的编码实践.